Light Bulbs in a circuit?

Question

There are three lightbulbs and are set up like this:
** _____lightbulb a______________
__|__*************** | **************|
* __**************bulb b****lightbulb c
***|_______________|___________|

What happens to the brightness of the remaining bulbs if A was removed and replaced by a wire? B was removed? If B was replaced by a wire.

Thank you all for helping me understand this! A medical student aren't smart in this stuff!

sorry the drawing is messed up. it supposed to be a retangle with A on the top and C on the right side and B in the middle ... the middle of the rectangle contains a line from and down. and there's a battery on the left side... HELP AND SORRY!

Answer 1

I am assuming that b & c are in parellel while a is in series with b&c. Circuit in series, if you remove a, then b & c will be brighter. but after you removed both a & c and replace them with wires, then c will be short circuit; therefore, c will not light. Hope this help!

Answer 2

EASY! I did that experiment. heres how it goes; if three lites are connected is(series) one after another, the first bulb would be the brightest. because as the power from the batt. is flowing through the bulbs it is used or lessoned by lighting each bulb. #1 is bright #2 not as bright #3 the dimmest. now as you replace each bulb with a piece of wire the energy from the battery is not lessoned and flows right through too bulb #2 or#3 get it!

Answer 3

If I understand your diagram correctly, then replacing bulb A with a wire should mean more current for B & C. Think about it -- bulbs consume power. One less bulb should mean more power for the other two bulbs "downstream"..

If you replace bulb B with a wire, then C should go dark, since electrons follow the path of least resistance and would naturally go around the resistive bulb if they could. Bulb A should become brighter, though, since there is no resistance downstream to slow down the current.